If N is the product of all multiples of 10 between 199 and 301, what i

Question

If N is the product of all multiples of 10 between 199 and 301, what is the greatest integer m for which  \frac{N}{10^m}  is an integer?

A. 10
B. 11
C. 13
D. 14
E. 15

globaldesi · Answer

the number of zeros in the product will be
list is 
200, 210, 220 .... , 300
so 
200 and 300 will give  10^2  * 10^2  = 10^4
and then 210 , 220, ...290
will give
10*10..10 (9 times) =  10^9 
However in product we have 2*5 also that will give 1 more zero
thus 10^1
thus total zeros =  10^{4+9+1}  =  10 ^{14} 
hence highest value of m will be 14 that could divide
IMO D

Stratos35 · Answer

globaldesi

I think the product has 2 more 2*5 because of 25=5*5 which potentially gives 2 pairs of 2*5 combined with 2 random 2s lets say from 26=2*13 and 22=2*11

In particular, Product = 200 * 210 * 220 *......*300 = 10*20 * 10*21 * 10*22 ..... 10*30 = 10^11 ( 20*21*22....*30)= 10^11( 2*10 *21* 2 *11*23*24* 5*5*2 *13*27*28*29*3* 10 )

So now we have 10^11 and 10 from 20, 10 from 30 and 2 2*5s from 25

globaldesi · Answer

I think I have taken one 5 already from 25 only 1 more remain that can be used with 24 to give one more 0 . I have already considered one 5 from it already
i think its 15

Stratos35 · Answer

globaldesi Yes exactly that was what I meant, the correct is 15 (E), you took the 1 2*5 and there is 1 more

shristi19 · Answer

To find the value of m , we have to find the no of 5 in N.
N= 200*210*220*230*240*250*260*270*280*290*300
No of 5 in each number = 2+1+1+1+1+3+1+1+1+1+2
Total no. of 5 in the product is 15,
so m is 15

astiles67 · Answer

Can someone explain what the concept here is? Looked in number properties guide already but could not figure out what rule is being applied. Specifically, why are we counting 10's and 5's as additional exponents for M? Why do we ignore the numbers that cant be divided by 5 or 10?

shaeazreem · Answer

Count how many of the multiples of 10 have more than one 5.

200
=
5
2
×
8

250
=
5
2
×
10
=
5
3
×
2

300
=
5
2
×
12

One 5 of each of these is counted in our 11 multiples of 10, but there are 4 additional 5s, which can be used to make 4 additional 10s. All in all, looking at the equation 
11
+
4
=
15
, there are fifteen 10s in 
N
. Thus, the greatest possible value of 
m
is 15.

pierjoejoe · Answer

the product is 
200*210*220...* 300

we have to find the greatest number m such that N/10^m is a integer

or N/(5*2)^m is a integers

we surely have the right amount of 2 in N, because they are common. we have to find out how many 5s are in the product. because if in the numerator we have less 5s with respect to the denominator than N/10^m will not be a integer.

we have to check how many numbers have 5 as factors in N. each one of them is a multiple of 10, so multiple of 5*2, so all of them are multiple of 5. we have:
(300-200)/10 +1 = 11 multiples of 10

we have to consider that 300 and 200 are 2 times multiples of 10 (and thus multiples of 5). the total is 11 + 2 = 13

250 is 2 * 5^3 --> 2*5 is the 10 we already counted in the initial 11

the other 2 5s in 250 will form a 10 with other 2s in the product N, thus we have to add them to the total:
(from the multiples of 10) + (from 200 and 300) + (from 250)
11 + 2 +2 = 15

bumpbot · Answer

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If N is the product of all multiples of 10 between 199 and 301, what i

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